Question

A rocket blast off with an acceleration of 5m/s^2. Its booster last for 35s. if then shuts off
How high did the rocket reach?

Answer 1

how did you get that

Answer 2

Well...not quite. Assuming the net acceleration of 5m/s^2:

Use the kinematic equation:
\[d=v_it+\frac{1}{2}at^2\]Since the initial velocity is 0m/s you can simplify that to:\[d=\frac{1}{2}at^2\]Now plug in the values and see what you get...

Answer 3

would it be 30.625m

Answer 4

jk

Answer 5

Check your math :)

Answer 6

If you are dealing with projections being shot upward or across you always need to take gravity into consideration. Well actually there is no acc. in the x direction only the y direction.

Answer 7

3062.5m

Answer 8

\[d=\frac{1}{2}(5m/s^2)(35s)^2=3062.5m\]If you factored in gravity...your rocket wouldn't be taking off at all...so the problem is assuming that you're actually moving upward at 5m/s^2.

Answer 9

what if you weren't factoring in gravity then what would the answer be

Answer 10

The one I posted :) I assumed a NET upward acceleration of 5m/s^2

Answer 11

3062.5m

Answer 12

so if you factored in gravity would the answer change

Answer 13

If you were to count gravity, your equation would look like this:\[d=\frac{1}{2}(5m/s^2-9.8m/s^2)(35s^2)\]That will give you a negative answer which means the rocket won't be taking off.

Answer 14

That's why I said that the the problem must be assuming that it's a NET upward acceleration of 5m/s^2....because otherwise the answer would not make sense.

Answer 15

changing the sign of gravity wouldn't make the answer logical?

Answer 16

If you took gravity to be + and upward to be - you would still obtain an answer with a - number?

Answer 17

oh ok

Answer 18

No...but if you did that, then you're assuming upward to be a negative direction which would result in a valid answer. Think about that for a moment :)

Answer 19

ok thank you

Answer 20

Maybe my last post isn't very clear. What I mean to say is that if you reverse both signs and end up with a negative answer, it would be valid since negative would be an upward movement. In that case, if your answer was positive, it would be invalid.

Answer 21

@annej> you're welcome :)

Answer 22

I swear, when I was using the kinematic equations in phy I would have to always use a positive g. Because the g in the formula was already turned into a + g from a - g. Lol

Answer 23

The sign of g just depends on how you set the problem up. It doesn't really matter as long as long as you stick with the signs for the directions throughout the problem.

I can walk through this one real quick and flip the signs...and you'll see that it won't matter.

Answer 24

Lose your sign...lose your mind, I always say :)

Answer 25

Haha okay I believe you.=)